A charge of 40 µC is pushed by a force of 15 µN a distance of 1 point 3 millimeters in an electric field. What is the electric potential difference?

The work done is force times distance: (15 × 10⁻⁶ N)(1.3 × 10⁻³ m) = 1.95 × 10⁻⁸ J. The potential difference is work divided by charge: (1.95 × 10⁻⁸ J)/(40 × 10⁻⁶ C) ≈ 4.9 × 10⁻⁴ V. This matches choice B, 4.9 × 10⁻⁴ V.

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